To determine at what temperatures the reaction is spontaneous, we use the Gibbs free energy change equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
For a reaction to be spontaneous, [tex]\(\Delta G\)[/tex] must be negative ([tex]\(\Delta G < 0\)[/tex]). Given the enthalpy change ([tex]\(\Delta H\)[/tex]) and the entropy change ([tex]\(\Delta S\)[/tex]), we need to find the temperatures at which:
[tex]\[ \Delta H - T \Delta S < 0 \][/tex]
Given values:
[tex]\[
\Delta H = -75 \text{ kJ/mol}
\][/tex]
[tex]\[
\Delta S = -0.081 \text{ kJ/(K·mol)}
\][/tex]
Substitute the given values into the inequality:
[tex]\[
-75 \text{ kJ/mol} - T (-0.081 \text{ kJ/(K·mol)}) < 0
\][/tex]
Simplify the inequality:
[tex]\[
-75 + 0.081T < 0
\][/tex]
Now, solve for [tex]\( T \)[/tex]:
[tex]\[
0.081T < 75
\][/tex]
[tex]\[
T < \frac{75}{0.081}
\][/tex]
[tex]\[
T < 925.9259259259259 \text{ K}
\][/tex]
Therefore, the reaction is spontaneous at temperatures [tex]\( T < 925.93 \text{ K} \)[/tex].
Hence, the correct answer is:
D. [tex]\(T < 930 \text{ K}\)[/tex]
